SOLUTION: 1. Solve algebraically the equation 15sin2x°=10cosx° for 0≤x<360
Can you provide full working.
Andy.
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Question 388619: 1. Solve algebraically the equation 15sin2x°=10cosx° for 0≤x<360
Can you provide full working.
Andy.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
15 sin 2x = 10 cos x. By double-angle formula, sin 2x = 2 sin x cos x, so
30 sin x cos x = 10 cos x. Move 10 cos x to the other side and divide by 10
3 sin x cos x - cos x = 0
cos x (3 sin x - 1) = 0
cos x = 0 or sin x = 1/3
x = 90, 270, or x = arcsin (1/3)
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